Question

A factory is producing iron bars. The length of the bars is believed to approximate a normal distribution. A simple random sample of 29 bars is selected and found to have an average length of 60.25 inches with a sample variance of .0625 inches.

a. What is the 95% confidence interval of the true average length of the iron bars?

b. Based on the result above, could the manufacturer claim the bars the factory produces iron bars 60 inches long? Explain your decision.

c. The manufacturer decided he wanted to redo the sample. They believe the shortest bars are about 59.75 inches and the longest about 60.25 inches. What should the sample size be for 95% confidence interval with a width of .05 inches?

Show all work please

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 29- 1 ) = 2.048
60.25 ± t(0.05/2, 29 -1) * 0.25/√(29)
Lower Limit = 60.25 - t(0.05/2, 29 -1) 0.25/√(29)
Lower Limit = 60.1549
Upper Limit = 60.25 + t(0.05/2, 29 -1) 0.25/√(29)
Upper Limit = 60.3451
95% Confidence interval is ( 60.1549 , 60.3451 )

Part b)

Values of confidence interval is greater than 60, hence we unable to conclude that  bars the factory produces iron bars 60 inches long

Part c)

Sample size can be calculated by below formula
n = ( 2 * Z(α/2) * σ / W )2
n = (2 * Z(0.05/2) * (0.25 / 0.05) )2
Critical value Z(α/2) = Z(α/2) = 1.96
n = (( 2 * 1.96 * (0.25 / 0.05 ) )2
n = 385
Required sample size at 95% confident is 385.